Any coordinate system in GR?
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From: Sorcerer on 29 Aug 2006 14:51 <carlip-nospam(a)physics.ucdavis.edu> wrote in message news:ed1v3g$mm$1(a)skeeter.ucdavis.edu... In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: > These two gentlemen fail to understand what the metric actually > represents. It is thus necessary to explain what the metric really > represents. Given the ubiquitously familiar Schwarzschild metric as > described in the very second equation, dt, dr, and dO (longitudinal > and latitudinal information) are the observed quantities. No, they aren't. Yes, they have. http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF Androcles
From: Koobee Wublee on 30 Aug 2006 16:21 carlip-nospam(a)physics.ucdavis.edu wrote: > In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: > ... As a simple example, compare the Schwarzschild > metric in isotropic coordinates and Schwarzschild coordinates (see > http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=node9.html > for the explicit expressions). Both describe the same spacetime, and > both have a radial coordinate called r. But at a given point, the > numerical value of r_Schwazrschild will be different from the value of > r_isotropic. I read one equation uses linear r, and the other one uses a silly non-linear radial measurement call r'. It is similar to a log scale for r, but in this case only someone got his head lost in the very curvature of space would embrace that time of scale system. The term r Schwarzschild is misleading. It should be called linear scale. Examp1e of this silly scale system is the following using the unit of meter primed, or m'. ** 1 m' = 10 m ** 2 m' = 5 m ** 3 m' = 4 m ** 4 m' = 4 m ** 5 m' = 5 m > Clearly they can't both be "the observed quantities." So how do you > propose to decide which one is "right"? (Note: an answer of the form, > "I saw the Schwarzschild form first" is not passing.) You are very correct that in this case on the choice of scale system a corresponding metric is appropriate for a particular scale system. However, there exists a system with also very linear scale that would render the new metric a very different independent metric. Both of these metrics when measured in linear scale are independent solutions to Einstein Field Equations in vacuum. Have you still not realized this? > For a *physical* example of the difference, see Bodenner and Will, > "Deflection of light to second order: A tool for illustrating principles > of general relativity," American Journal of Physics, Vol. 71, No. 8, > pp. 770-773, August 2003. You will learn that the coordinate values > for the deflection of light by the Sun differ depending on which "r" > you use, and that the correct result can only be obtained by using > "measurable, coordinate-independent quantities." Yes, however, what I am interested in the linear scale. As an observer, I want to use my own coordinate system that tells me it is a linear scale. >> With the following spacetime, >> >> ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 / >> r^2) (1 + r / K)^2 dO^2 >> >> How do you specify a distance 2 AU from the center of the sun? > > First of all, you can't use this metric to answer that question -- > the Sun is not a vacuum, and the metric you've written down is only > valid in empty space. To answer the question as stated, you need > first to give an interior metric that "glues" on to the exterior > metric you've written down. The answer will depend on the details > of that interior metric. I am sorry. I meant (AU = astronomical unit) where (2 AU >> radius of the sun). Notice I am using the linear scale here. > You can, however, answer the question, "How do you specify a distance > 2 AU from the edge of the Sun." To answer that, you need to take > the interval ds at fixed time and fixed angle (so dt=0 and dO=0) and > integrate ds from the edge of the Sun to an arbitrary value r. That > integral will give you the physical ("proper") distance D, as measured > by an observer at rest. Inverting the answer, you can find r as a > function of D. Take that expression and set D = 2 AU. I thought about this a few months ago. This does not appeal to me. The reason is that the metric must be observed to be the same from one observer to another. Since each observer cannot directly observe a curvature in space, he observes a curvature in space as if it is normal. Even if the radius component is curved, an observer will always observe the surface area to be (4 pi R^2) with (R = radius). Thus, the term (1 - 2 G M / r / c^2) has (r = 0) right at the center of M. > This is fairly basic general relativity. It's true that beginners > sometimes get coordinate differences and physical quantities mixed up, > but if you take a decent course in the subject, or actually read a > decent book (and work through the exercises, etc.), you should be able > to correct that kind of mistake pretty quickly. Actually, this is fairly basic concept of curvature in spacetime in which GR is just a special case justifying a specific existence of a metric. > (For example, see the discussion on page 191 of Weinberg's text about > the problem of whether a prediction "really refers to an objective > physical measurement or whether it has folded into it arbitrary > subjective elements dependent on our choice of coordinate system"; > or, for the meaning of Schwarzschild coordinates, Wald's careful > discussion in section 6.1 of his text. And by all means, read the > paper by Bodenner and Will that I cited above, which is about almost > precisely this issue.) I would sometimes use log scale. However, under all other circumstances, linear scale is the way to go. You can return all these weird scaling systems to their rightful owners in curved space. Thank you very much.
From: JanPB on 31 Aug 2006 00:59 Koobee Wublee wrote: > carlip-nospam(a)physics.ucdavis.edu wrote: > > In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: > > > ... As a simple example, compare the Schwarzschild > > metric in isotropic coordinates and Schwarzschild coordinates (see > > http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=node9.html > > for the explicit expressions). Both describe the same spacetime, and > > both have a radial coordinate called r. But at a given point, the > > numerical value of r_Schwazrschild will be different from the value of > > r_isotropic. > > I read one equation uses linear r, and the other one uses a silly > non-linear radial measurement call r'. It is similar to a log scale > for r, but in this case only someone got his head lost in the very > curvature of space would embrace that time of scale system. The term r > Schwarzschild is misleading. It should be called linear scale. > > Examp1e of this silly scale system is the following using the unit of > meter primed, or m'. > > ** 1 m' = 10 m > ** 2 m' = 5 m > ** 3 m' = 4 m > ** 4 m' = 4 m > ** 5 m' = 5 m Coordinates are non-physical. They are man-made labels. Why do you think it matters how events are labelled? -- Jan Bielawski
From: carlip-nospam on 31 Aug 2006 17:37 In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: > carlip-nospam(a)physics.ucdavis.edu wrote: >> In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: >> ... As a simple example, compare the Schwarzschild >> metric in isotropic coordinates and Schwarzschild coordinates (see >> http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=node9.html >> for the explicit expressions). Both describe the same spacetime, and >> both have a radial coordinate called r. But at a given point, the >> numerical value of r_Schwazrschild will be different from the value of >> r_isotropic. > I read one equation uses linear r, and the other one uses a silly > non-linear radial measurement call r'. As you've written this, it doesn't actually mean anything. Linearity is a relationship between two quantities. When you say "linear r," what do you mean? Linear in what? My guess -- though it is only a guess -- is that you mean "linear in physical distance." If so, you are dodging the question: *why* do you think that one r, rather than the other, is linear in distance? What is your evidence for this? > It is similar to a log scale > for r, but in this case only someone got his head lost in the very > curvature of space would embrace that time of scale system. The term r > Schwarzschild is misleading. It should be called linear scale. Really? Why is the Schwarzschild r, and not, say, the isotropic r, "linear"? (I would have thought you would prefer the isotropic coordinates. After all, if you write r^2=x^2+y^2+z^2, the spatial metric in isotropic coordinates is of the form F(r)(dx^2+dy^2+dz^2), while the metric in Schwarzschild coordinates has nonorthogonal pieces like (xy/r^2)dxdy.) [...] >> For a *physical* example of the difference, see Bodenner and Will, >> "Deflection of light to second order: A tool for illustrating principles >> of general relativity," American Journal of Physics, Vol. 71, No. 8, >> pp. 770-773, August 2003. You will learn that the coordinate values >> for the deflection of light by the Sun differ depending on which "r" >> you use, and that the correct result can only be obtained by using >> "measurable, coordinate-independent quantities." > Yes, however, what I am interested in the linear scale. As an > observer, I want to use my own coordinate system that tells me > it is a linear scale. How does a coordinate system "tell you" that? If I didn't have you to report what you appear to consider a revealed truth, what *specific* property of Schwarzschild coordinates could I use to determine that the Schwarzschild r and not the isotropic r is "linear"? Steve Carlip
From: Koobee Wublee on 31 Aug 2006 21:20
carlip-nospam(a)physics.ucdavis.edu wrote: > In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: > > carlip-nospam(a)physics.ucdavis.edu wrote: > >> In sci.physics Koobee Wublee <koobee.wublee(a)gmail.com> wrote: > > >> ... As a simple example, compare the Schwarzschild > >> metric in isotropic coordinates and Schwarzschild coordinates (see > >> http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=node9.html > >> for the explicit expressions). Both describe the same spacetime, and > >> both have a radial coordinate called r. But at a given point, the > >> numerical value of r_Schwazrschild will be different from the value of > >> r_isotropic. > > > I read one equation uses linear r, and the other one uses a silly > > non-linear radial measurement call r'. > > As you've written this, it doesn't actually mean anything. Linearity > is a relationship between two quantities. When you say "linear r," > what do you mean? Linear in what? What I mean linear is that the unit of measure for distance is in a constant multiple of the meter. > My guess -- though it is only a guess -- is that you mean "linear in > physical distance." If so, you are dodging the question: *why* do > you think that one r, rather than the other, is linear in distance? > What is your evidence for this? I still don't understand why you would think this subject is so difficult to understand. > > It is similar to a log scale > > for r, but in this case only someone got his head lost in the very > > curvature of space would embrace that time of scale system. The term r > > Schwarzschild is misleading. It should be called linear scale. > > Really? Why is the Schwarzschild r, and not, say, the isotropic r, > "linear"? OK, what do you mean by Schwarzschild r? What unit of measure do you use? > (I would have thought you would prefer the isotropic coordinates. > After all, if you write r^2=x^2+y^2+z^2, the spatial metric in > isotropic coordinates is of the form F(r)(dx^2+dy^2+dz^2), while > the metric in Schwarzschild coordinates has nonorthogonal pieces > like (xy/r^2)dxdy.) I like to use meters are the choice off unit measurement. > [...] > >> For a *physical* example of the difference, see Bodenner and Will, > >> "Deflection of light to second order: A tool for illustrating principles > >> of general relativity," American Journal of Physics, Vol. 71, No. 8, > >> pp. 770-773, August 2003. You will learn that the coordinate values > >> for the deflection of light by the Sun differ depending on which "r" > >> you use, and that the correct result can only be obtained by using > >> "measurable, coordinate-independent quantities." > > > Yes, however, what I am interested in the linear scale. As an > > observer, I want to use my own coordinate system that tells me > > it is a linear scale. > > How does a coordinate system "tell you" that? If I didn't have you to > report what you appear to consider a revealed truth, what *specific* > property of Schwarzschild coordinates could I use to determine that > the Schwarzschild r and not the isotropic r is "linear"? In both metrics, I can use my choice of unit measurement that is the meter. Thus, both metrics are valid solutions to Einstein Field Equations in vacuum, and there are an infinite numbers of solutions out there including one that models the following. This is achieved without invoking the negative energy density in free space known as the Cosmological Constant. ** Newtonian law of gravity at short distances ** Constant expanding universe at medium distances ** Accelerated expanding universe at farther distances The spacetime with this metric is ds^2 = c^2 A(r) dt^2 - B(r) dr^2 - C(r) dO^2 Where ** A(r) = 1 / (1 + K / r + r / L + r^2 / (P1 N)) ** B(r) = (1 + K / r + r / L + r^2 / (L N)) (1 - r^2 / (K L) - 2 r^3 / (K L N))^2 / (1 + r^2 / (K L) + r^3 / (K L N))^4 ** C(r) = r^2 (1 + K / r + r / P1 + r^2 / (L N))^2 / (1 + r^2 / (K L) + r^3 / (K L N))^2 ** K = 2 G M / c^2 ** L, N = Constants in cosmic scale ** K / r = Model of Newtonian gravity ** r / L = Model of constant expanding universe ** r^2 / N = Model of accelerated expanding universe Do you choose not to accept this point because of your deep-rooted faith in GR? You can choose to believe in the emperor's clothes for all you want, but the next generations of students are going to be exposed to what I have pointed out. They will not have this faith to burden themselves down. They will see that the emperor indeed has no clothes on. |